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A cylindrical disc, in the context of the spinal column, refers to the thin, disk-shaped cartilage located between each pair of vertebrae. This cartilage is essential because it acts as a cushion, allowing for movement and absorbing impacts between the bones of the spine. The disc's radius and thickness are key dimensions that are critical to calculations involving deformation under applied forces. In this specific exercise, the disc has a radius of about 0.03 meters and a thickness of about 0.007 meters.

These dimensions allow us to determine the surface area of the top face of the disc, which is crucial when considering how forces affect the disc. The formula for the area of a circle, \( A = \pi r^2 \), gives us the surface area when plugging in the radius. Understanding this basic geometry helps in calculating shear forces exerted on the disc.

Shear strain is a measure of how much a material deforms in response to a shearing force. It is a dimensionless number that defines the extent of deformation in a material when subjected to stress parallel to its face.

In the exercise, shear strain is represented with the symbol \( \gamma \), and it quantifies the change in shape of the disc's material due to the applied shearing force. The formula \( \gamma = \frac{F}{A \, G} \) connects shear strain to several key elements: the shearing force \( F \), the area \( A \) over which the force is applied, and the shear modulus \( G \) of the material. This relationship helps us intuitively understand how changing any of these variables affects the deformation of the disc.

Shear stress is another crucial concept and refers to the internal force that the material resists deformation by. It is defined as the force per unit area and is measured in pascals (Pa).

In the given problem, a force of 11 N is applied parallel to the top surface of the disc. As a result, shear stress becomes a critical part of the analysis. The relationship between shear stress and shear strain, given by the formula \( \text{shear stress} = \text{shear modulus} \times \text{shear strain} \), allows us to understand how the material reacts internally to the exerted force. This concept aids in predicting how force application affects the structure and calculation of deformation.

Displacement calculation is the final step in understanding how an external force changes the position of the top surface of the cylindrical disc in relation to its fixed bottom surface. As shear strain is defined as \( \gamma = \frac{\Delta x}{h} \), where \( \Delta x \) is the relative displacement and \( h \) is the thickness of the disc, rearranging this formula allows us to find the displacement.

This entails substituting the known values of shear strain and the disc's thickness into the equation \( \Delta x = \gamma \times h \). Through this substitution, we see how minimal changes in these parameters can greatly affect the calculated displacement. In the scenario from the exercise, understanding these relationships through displacement calculation confirms the degree of deformation, ensuring a comprehensive grasp of how materials behave under shear forces.